Sketch the region enclosed by x+y^2=42 and x+y=0 . Decide whether to integrate w

Question

Explanation / Answer

x + y^2 = 42 --> [y=-sqrt(42-x),y=sqrt(42-x)] x + y = 0 -----> y=-x A window of [x,-8,43],[y,-8,8] will show you all relevant detail. You will see a parabola sideways, open to the left, being crossed by a straight line, having negative slope. The x-axis in the middle divides the parabola into two equal halves, one upper and one lower. The enclosed region then is the area of the parabola up to the straight line y=-x. The straight line intersects the parabola at P1(6,-6) and P2 (-7,7). You find the y-values by equating: y^2+x-42=y+x : [y=-6,y=7] You find the x-values by equating: -6=-sqrt(42-x): [x=6] and 7=sqrt(42-x) : [x=-7]. I decided to integrate with respect to y So I solved both equations for x: I) x=42-y^2 II) x=-y The total area of the parabola from y=-7 to y=7 is integrate(42-y^2,y,-7,7)=1078/3 The wedge to subtract is: integrate(42-y^2,y,-7,-6)+ abs(integrate(-y,y,-6,0))+ abs(integrate(-y,y,0,7))=253/6 Hence 1078/3-253/6=317.1666

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.